How to prove the ratio $ \dfrac{x(2017x + y)}{2018x + y} $ in the simplest form?

Question

Let $x, y$ be positive integers such that $(x, y) = 1.$ Prove that the ratio $$ \dfrac{x(2017x + y)}{2018x + y} $$ in the simplest form.

Answer 1

Assume that there exists a natural number $n\ge 2$ so that $$ \begin{cases} n|x(2017x+y) \\ n|(2018x+y) \end{cases}$$

Then, there also exists a prime number $d\ge 2$ ($d$ is a factor of $n$) such that

$$ \begin{cases} d|x(2017x+y) \\ d|(2018x+y) \end{cases}$$

Then we will have \begin{cases} d|x(2017x+y) \\ d|x+2017x+y \end{cases}

Bacause $d$ is a prime number, either $d|x$ or $d|2017x+y$ (first expression), from the second expression we will have:

• $d|x\Rightarrow \begin{cases}d|2017x+y \\ d|2017x \end{cases} \Rightarrow \begin{cases}d|x\\ d|y \end{cases}\text{ (contradiction)}$
• $d|2017x+y\Rightarrow \begin{cases}d|x \\ d|2017x \end{cases}\Rightarrow \begin{cases}d|x\\ d|y \end{cases}\text{ (contradiction)}$